Last 2 Digits | Divisibility | Mockat

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Last 2 Digits

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Module 6 of 8

Overview

4.2 Last two digits

Questions around last two digits are less common in entrance tests. However, the concept is quite simple and easy to learn. The cases below are for finding the last two digits of a number

x^n

where

x

and

n

are integers.

4.2.1 x is a multiple of $\bm{5}$

Where

n > 1

, last digit of

x

5

and tens digit of x is even, the last two digits of

\bm{x^{n} = 25}

.

Where

n > 1

, last digit of

x

5

and tens digit of x is odd,
1) the last two digits of

x^{n} = 25

, where

n

is even; and
2) the last two digits of

x^{n} = 75

, where

n

is odd.

4.2.2 x is a multiple of $\bm{10}$

Where

n > 1

and last digit of

x

0

, the last two digits of

\bm{x^n = 00}.

4.2.3 x is a multiple of $\bm{2}$ but not $\bm{4}$

Where

k

is an integer, last two digits of

\bm{x^{40k +1}} \space =

last two digits of

\bm{x + 50}

.

We, therefore, split the powers and then individually multiply, as shown in the example below:

Example 18

What are the last two digits of

86^{283}

Solution

86

is a multiple of

2

but not

4

.

∴

T(86^{281}) = T(86^{40 \times 7 + 1}) = T(86 + 50) = 36

T(86^{283}) = T(86^{281} \times 86^{2}) = T(36 \times 96) = T(3456) = \bm{56}

Answer:

56

4.2.4 x is a multiple of $\bm{4}$

Where

k

is an integer, last two digits of

\bm{x^{40k + 1}} =

last two digits of

\bm{x}

Example 19

What are the last two digits of

2016^{2016}

Solution

We need to find

T(16^{2016})

16

is a multiple of

4

.

∴

T(16^{40 \times 50 + 1}) = T(16^{2001}) = 16

T(16^{2016}) = T(16^{2001} \times 16^{15}) = T(16 \times 16^{15}) = \bm{T(16^{16})}

After this, we normally multiply the numbers to find the last two digits.

T(16^2) = 56

T(16^4) = T(56^2) = 36

T(16^3) = T(36^2) = 96

T(16^{16}) = T(96^2) = 16

Answer:

16

4.2.5 x is an odd number other than $\bm{5}$

When the units digit of a number is

1

, then there is a cyclicity in the tens digit. Using this pattern, where the last digit of

x

1

3

7

9

1

) Raise

x

to a power such that the units digit becomes

1

2

) The units digit of the resulting number raised to any power will always be

1

3

) The tens digit of this number is the product of the tens digit of the base and the units digit of the power.
From steps

2

and

3

, if

x = \bm{\underline{7}1^{48\underline{6}}}

Units digit of

x = 1

Tens digit of

x = U(7 \times 6) = 2

Last

2

digits of

71^{486} = 21

Example 20

What are the last two digits of

637^{92}

Solution

T(637^{92}) = T \left((37^{2})^{46} \right)

= T(69^{46}) = T\left((69^{2})^{23} \right) = T(61^{23})

Units digit of

61^{23} = 1

Tens digit of

61^{23} = U(6 \times 3) = 8

∴

T(61^{23}) = \bm{81}

Answer:

81

CAT 2025 Lesson : Divisibility - Last 2 Digits

Divisibility

4.2 Last two digits

4.2.1 x is a multiple of 5\bm{5}5

4.2.2 x is a multiple of 10\bm{10}10

4.2.3 x is a multiple of 2\bm{2}2 but not 4\bm{4}4

Example 18

Solution

4.2.4 x is a multiple of 4\bm{4}4

Example 19

Solution

4.2.5 x is an odd number other than 5\bm{5}5

Example 20

Solution

4.2.1 x is a multiple of $\bm{5}$

4.2.2 x is a multiple of $\bm{10}$

4.2.3 x is a multiple of $\bm{2}$ but not $\bm{4}$

4.2.4 x is a multiple of $\bm{4}$

4.2.5 x is an odd number other than $\bm{5}$